Simplify the following expression and state the condition under which the simplification is valid. You can assume that $t \neq 0$. $z = \dfrac{2t}{14t + 7} \times \dfrac{6t + 3}{-6} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $z = \dfrac{ 2t \times (6t + 3) } { (14t + 7) \times -6 } $ $ z = \dfrac {2t \times 3(2t + 1)} {-6 \times 7(2t + 1)} $ $ z = \dfrac{6t(2t + 1)}{-42(2t + 1)} $ We can cancel the $2t + 1$ so long as $2t + 1 \neq 0$ Therefore $t \neq -\dfrac{1}{2}$ $z = \dfrac{6t \cancel{(2t + 1})}{-42 \cancel{(2t + 1)}} = -\dfrac{6t}{42} = -\dfrac{t}{7} $